The apparent coefficients of volume expansion of a liquid when heated filled in vessel A and B of identical volumes, are found to be `gamma_(1)` and `gamma_(2)` respectively. If `alpha_(1)` be the coefficient of liner expansion for A, then what will be the be the coefficient of linear expansion for B? (True expansion - vessel expansion = app.exp)

Options

A. `((gamma_(1)-gamma_(2)))/(3)-alpha_(1)`
B. `((gamma_(2)-gamma_(1)))/(3)+alpha_(1)`
C. `((gamma_(2)-gamma_(1)))/(3)-alpha_(1)`
D. `((gamma_(1)-gamma_(2)))/(3)+alpha_(1)`

Answer

To solve the problem, we need to find the coefficient of linear expansion for vessel B (denoted as \( \alpha_2 \)) given the apparent coefficients of volume expansion \( \gamma_1 \) and \( \gamma_2 \) for vessels A and B, respectively, and the coefficient of linear expansion for vessel A (\( \alpha_1 \)).

### Step-by-Step Solution:

1. **Understanding Apparent Volume Expansion**:
The apparent volume expansion of a liquid in a vessel can be expressed as:
\[ \text{True Expansion} - \text{Vessel Expansion} = \text{Apparent Expansion} \]
This means:
\[ \Delta V_L - \Delta V_C = \Delta V_{apparent} \]
where \( \Delta V_L \) is the true expansion of the liquid, and \( \Delta V_C \) is the expansion of the vessel.

2. **Expressing True and Vessel Expansion**:
The true volume expansion of the liquid can be expressed as:
\[ \Delta V_L = V \cdot \gamma_L \cdot \Delta T \]
The volume expansion of the vessel can be expressed as:
\[ \Delta V_C = V \cdot \gamma_C \cdot \Delta T \]
where \( V \) is the initial volume of the liquid.

3. **Setting Up the Equation for Vessel A**:
For vessel A, we have:
\[ \Delta V_L - \Delta V_C = V \cdot \gamma_1 \cdot \Delta T \]
Substituting the expansions:
\[ V \cdot \gamma_L - V \cdot \gamma_C = V \cdot \gamma_1 \]
Dividing through by \( V \):
\[ \gamma_L - \gamma_C = \gamma_1 \]
Rearranging gives:
\[ \gamma_L = \gamma_1 + \gamma_C \tag{1} \]

4. **Setting Up the Equation for Vessel B**:
Similarly, for vessel B, we have:
\[ \Delta V_L - \Delta V_C = V \cdot \gamma_2 \cdot \Delta T \]
Substituting the expansions:
\[ V \cdot \gamma_L - V \cdot \gamma_C = V \cdot \gamma_2 \]
Dividing through by \( V \):
\[ \gamma_L - \gamma_C = \gamma_2 \]
Rearranging gives:
\[ \gamma_L = \gamma_2 + \gamma_C \tag{2} \]

5. **Equating the Two Expressions**:
From equations (1) and (2), we can equate the two expressions for \( \gamma_L \):
\[ \gamma_1 + \gamma_C = \gamma_2 + \gamma_C \]
This simplifies to:
\[ \gamma_1 = \gamma_2 \]

6. **Relating Coefficients of Linear Expansion**:
The relationship between the coefficients of volume expansion (\( \gamma \)) and linear expansion (\( \alpha \)) is given by:
\[ \gamma = 3\alpha \]
Therefore, we can express the coefficients for vessels A and B:
\[ \gamma_L = 3\alpha_1 \quad \text{(for vessel A)} \]
\[ \gamma_L = 3\alpha_2 \quad \text{(for vessel B)} \]

7. **Setting Up the Final Equation**:
Equating the two expressions for \( \gamma_L \):
\[ 3\alpha_1 + \gamma_C = 3\alpha_2 + \gamma_C \]
This simplifies to:
\[ 3\alpha_1 = 3\alpha_2 \]

8. **Solving for \( \alpha_2 \)**:
Thus, we find:
\[ \alpha_2 = \alpha_1 + \frac{\gamma_1 - \gamma_2}{3} \]

### Final Result:
The coefficient of linear expansion for vessel B is given by:
\[ \alpha_2 = \alpha_1 + \frac{\gamma_1 - \gamma_2}{3} \]